[curiouscat]

Just got thinking, is the Wurtz Reaction feasible with di-haloalkanes?

Say, with something like

X-R-X + M → X-R• + M⁺X⁻

X-R• + M → X-R−M⁺

X-R−M⁺ + X-R-X → X-R-R-X + M⁺X⁻

M=Na, say.

If so would the X-R-R-X again be available for further reaction and thus build up the chain? Or am I missing something obvious?

[Schrödinger]

Well, crudely, we kinda need to consider the stability of the radical (which later becomes a carbanion) formed right? So if you were to put a negative charge on the end of a chain, isn't it less stable compared to the one which has a halogen right beside it?

[curiouscat]

Well, crudely, we kinda need to consider the stability of the radical (which later becomes a carbanion) formed right? So if you were to put a negative charge on the end of a chain, isn't it less stable compared to the one which has a halogen right beside it?

Hmm...I'm not sure whether my mechanism's entirely correct; but if it is which step are talking about? Can you explain a bit more?

You think the reaction may be a total non-starter (not excluding that possibility!) or that subsequent growth to larger chains is unlikely?

[Schrödinger]

X-R• + M → X-R−M+

The product which you've written, is it not XR^-M⁺? That is, the negative charge is on the alkyl chain (Grignard reagent, for instance). The negative charge on the XR^- in the product of the above reaction is stable thanks to the electron withdrawing attitude of the halogen/other electron withdrawing groups. But consider X-R-R^-. The carbanion's neighbour is not an electron withdrawing group, but an alkyl group more or less. Hence the instability?

[curiouscat]

X-R• + M → X-R−M+

The product which you've written, is it not XR^-M⁺? That is, the negative charge is on the alkyl chain (Grignard reagent, for instance). The negative charge on the XR^- in the product of the above reaction is stable thanks to the electron withdrawing attitude of the halogen/other electron withdrawing groups. But consider X-R-R^-. The carbanion's neighbour is not an electron withdrawing group, but an alkyl group more or less. Hence the instability?

So is what you are saying equivalent to saying that Wurtz couldn't couple a long chain alkyl halide?

Would this reaction, say, happen or not?

[souro10]

Well, 1,3 dihaloalkanes to 1,6 dihaloalkanes form corresponding cyclic compounds under the conditions of Wurtz Reaction. In fact, this is one of the general methods of preparation of cycloalkanes- cyclopropane , cyclobutane, etc.

Maybe 1,2 dihaloalkanes (2 moles would be required ) may undergo coupling to form cyclic compounds?
For example 2 moles of the compound in the first post might react under conditions of wurtz reaction to form cyclobutane?

" X-R-X + M → X-R• + M+X−

X-R• + M → X-R−M+

X-R−M+ + X-R-X → X-R-R-X + M+X−

M=Na "

So at first it'll form Cl-CH₂-CH₂-CH₂-CH₂-Cl -- (1)

And now we have a 1,4 dihaloalkane, which can form cyclobutane under the reaction conditions?

I think stoichiometric ratio will play a very important role too..
May be under limited conditions of the metal, the reaction will stop at (1) and that'll be the final product. But if more metal is present, it may react all the way upto the cycloalkane, after which no further reaction is feasible.

[opsomath]

You'd run a very good chance of reducing this thing to ethylene as well.

[curiouscat]

You'd run a very good chance of reducing this thing to ethylene as well.

Could be. Actually what got me thinking about this was whether by throwing in some Na into dibromocubane could I get cubanes to couple into a chain?



Sigh, can't get the SMILES right to draw the chain I mean....

[Schrödinger]

I was actually referring to your reaction mixture, and was trying to invoke relative reactivities rather than absolute stability. There will be competition between the 2 kinds of carbanions and I just felt one might have the edge, being attached to a halogen.

[opsomath]

Well, that's certainly an interesting structure you propose. My guess is that the cubane monoradical would find some way to ring-open and release some of that crapton of strain.

Could be. Actually what got me thinking about this was whether by throwing in some Na into dibromocubane could I get cubanes to couple into a chain?



Sigh, can't get the SMILES right to draw the chain I mean....

[curiouscat]

Thanks. You might be right!

How unstable Cubanes, in general, because of the strain?